Calculus Examples

$\ln (3 x^{2} - 2 x + 1)$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$3 x^{2} - 2 x + 1 = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 1.2.2

Substitute the values $a = 3$ , $b = - 2$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (3 \cdot 1)}}{2 \cdot 3}$

Step 1.2.3

Simplify.

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Step 1.2.3.1

Simplify the numerator.

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Step 1.2.3.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}$

Step 1.2.3.1.2

Multiply $- 4 \cdot 3 \cdot 1$ .

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Step 1.2.3.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{2 \pm \sqrt{4 - 12 \cdot 1}}{2 \cdot 3}$

Step 1.2.3.1.2.2

Multiply $- 12$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 12}}{2 \cdot 3}$

Step 1.2.3.1.3

Subtract $12$ from $4$ .

$x = \frac{2 \pm \sqrt{- 8}}{2 \cdot 3}$

Step 1.2.3.1.4

Rewrite $- 8$ as $- 1 (8)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 8}}{2 \cdot 3}$

Step 1.2.3.1.5

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{8}}{2 \cdot 3}$

Step 1.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{8}}{2 \cdot 3}$

Step 1.2.3.1.7

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 1.2.3.1.7.1

Factor $4$ out of $8$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (2)}}{2 \cdot 3}$

Step 1.2.3.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 2}}{2 \cdot 3}$

Step 1.2.3.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{2})}{2 \cdot 3}$

Step 1.2.3.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{2}}{2 \cdot 3}$

Step 1.2.3.2

Multiply $2$ by $3$ .

$x = \frac{2 \pm 2 i \sqrt{2}}{6}$

Step 1.2.3.3

Simplify $\frac{2 \pm 2 i \sqrt{2}}{6}$ .

$x = \frac{1 \pm i \sqrt{2}}{3}$

Step 1.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.2.4.1

Simplify the numerator.

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Step 1.2.4.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}$

Step 1.2.4.1.2

Multiply $- 4 \cdot 3 \cdot 1$ .

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Step 1.2.4.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{2 \pm \sqrt{4 - 12 \cdot 1}}{2 \cdot 3}$

Step 1.2.4.1.2.2

Multiply $- 12$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 12}}{2 \cdot 3}$

Step 1.2.4.1.3

Subtract $12$ from $4$ .

$x = \frac{2 \pm \sqrt{- 8}}{2 \cdot 3}$

Step 1.2.4.1.4

Rewrite $- 8$ as $- 1 (8)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 8}}{2 \cdot 3}$

Step 1.2.4.1.5

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{8}}{2 \cdot 3}$

Step 1.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{8}}{2 \cdot 3}$

Step 1.2.4.1.7

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 1.2.4.1.7.1

Factor $4$ out of $8$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (2)}}{2 \cdot 3}$

Step 1.2.4.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 2}}{2 \cdot 3}$

Step 1.2.4.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{2})}{2 \cdot 3}$

Step 1.2.4.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{2}}{2 \cdot 3}$

Step 1.2.4.2

Multiply $2$ by $3$ .

$x = \frac{2 \pm 2 i \sqrt{2}}{6}$

Step 1.2.4.3

Simplify $\frac{2 \pm 2 i \sqrt{2}}{6}$ .

$x = \frac{1 \pm i \sqrt{2}}{3}$

Step 1.2.4.4

Change the $\pm$ to $+$ .

$x = \frac{1 + i \sqrt{2}}{3}$

Step 1.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.2.5.1

Simplify the numerator.

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Step 1.2.5.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}$

Step 1.2.5.1.2

Multiply $- 4 \cdot 3 \cdot 1$ .

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Step 1.2.5.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{2 \pm \sqrt{4 - 12 \cdot 1}}{2 \cdot 3}$

Step 1.2.5.1.2.2

Multiply $- 12$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 12}}{2 \cdot 3}$

Step 1.2.5.1.3

Subtract $12$ from $4$ .

$x = \frac{2 \pm \sqrt{- 8}}{2 \cdot 3}$

Step 1.2.5.1.4

Rewrite $- 8$ as $- 1 (8)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 8}}{2 \cdot 3}$

Step 1.2.5.1.5

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{8}}{2 \cdot 3}$

Step 1.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{8}}{2 \cdot 3}$

Step 1.2.5.1.7

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 1.2.5.1.7.1

Factor $4$ out of $8$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (2)}}{2 \cdot 3}$

Step 1.2.5.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 2}}{2 \cdot 3}$

Step 1.2.5.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{2})}{2 \cdot 3}$

Step 1.2.5.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{2}}{2 \cdot 3}$

Step 1.2.5.2

Multiply $2$ by $3$ .

$x = \frac{2 \pm 2 i \sqrt{2}}{6}$

Step 1.2.5.3

Simplify $\frac{2 \pm 2 i \sqrt{2}}{6}$ .

$x = \frac{1 \pm i \sqrt{2}}{3}$

Step 1.2.5.4

Change the $\pm$ to $-$ .

$x = \frac{1 - i \sqrt{2}}{3}$

Step 1.2.6

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{2}}{3}, \frac{1 - i \sqrt{2}}{3}$

Step 1.3

The vertical asymptote occurs at $x = \frac{1 + i \sqrt{2}}{3}, x = \frac{1 - i \sqrt{2}}{3}$ .

Vertical Asymptote: $x = \frac{1 + i \sqrt{2}}{3}, x = \frac{1 - i \sqrt{2}}{3}$

Step 2

Find the point at $x = 1$ .

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Step 2.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \ln (3 {(1)}^{2} - 2 \cdot 1 + 1)$

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

One to any power is one.

$f (1) = \ln (3 \cdot 1 - 2 \cdot 1 + 1)$

Step 2.2.1.2

Multiply $3$ by $1$ .

$f (1) = \ln (3 - 2 \cdot 1 + 1)$

Step 2.2.1.3

Multiply $- 2$ by $1$ .

$f (1) = \ln (3 - 2 + 1)$

Step 2.2.2

Simplify by adding and subtracting.

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Step 2.2.2.1

Subtract $2$ from $3$ .

$f (1) = \ln (1 + 1)$

Step 2.2.2.2

Add $1$ and $1$ .

$f (1) = \ln (2)$

Step 2.2.3

The final answer is $\ln (2)$ .

$\ln (2)$

Step 2.3

Convert $\ln (2)$ to decimal.

$y = 0.69314718$

Step 3

Find the point at $x = 2$ .

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Step 3.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = \ln (3 {(2)}^{2} - 2 \cdot 2 + 1)$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Raise $2$ to the power of $2$ .

$f (2) = \ln (3 \cdot 4 - 2 \cdot 2 + 1)$

Step 3.2.1.2

Multiply $3$ by $4$ .

$f (2) = \ln (12 - 2 \cdot 2 + 1)$

Step 3.2.1.3

Multiply $- 2$ by $2$ .

$f (2) = \ln (12 - 4 + 1)$

Step 3.2.2

Simplify by adding and subtracting.

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Step 3.2.2.1

Subtract $4$ from $12$ .

$f (2) = \ln (8 + 1)$

Step 3.2.2.2

Add $8$ and $1$ .

$f (2) = \ln (9)$

Step 3.2.3

The final answer is $\ln (9)$ .

$\ln (9)$

Step 3.3

Convert $\ln (9)$ to decimal.

$y = 2.19722457$

Step 4

Find the point at $x = 3$ .

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Step 4.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = \ln (3 {(3)}^{2} - 2 \cdot 3 + 1)$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $3$ by ${(3)}^{2}$ by adding the exponents.

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Step 4.2.1.1.1

Multiply $3$ by ${(3)}^{2}$ .

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Step 4.2.1.1.1.1

Raise $3$ to the power of $1$ .

$f (3) = \ln (3 {(3)}^{2} - 2 \cdot 3 + 1)$

Step 4.2.1.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (3) = \ln (3^{1 + 2} - 2 \cdot 3 + 1)$

Step 4.2.1.1.2

Add $1$ and $2$ .

$f (3) = \ln (3^{3} - 2 \cdot 3 + 1)$

Step 4.2.1.2

Raise $3$ to the power of $3$ .

$f (3) = \ln (27 - 2 \cdot 3 + 1)$

Step 4.2.1.3

Multiply $- 2$ by $3$ .

$f (3) = \ln (27 - 6 + 1)$

Step 4.2.2

Simplify by adding and subtracting.

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Step 4.2.2.1

Subtract $6$ from $27$ .

$f (3) = \ln (21 + 1)$

Step 4.2.2.2

Add $21$ and $1$ .

$f (3) = \ln (22)$

Step 4.2.3

The final answer is $\ln (22)$ .

$\ln (22)$

Step 4.3

Convert $\ln (22)$ to decimal.

$y = 3.09104245$

Step 5

The log function can be graphed using the vertical asymptote at $x = \frac{1 + i \sqrt{2}}{3}, x = \frac{1 - i \sqrt{2}}{3}$ and the points $(1, 0.69314718), (2, 2.19722457), (3, 3.09104245)$ .

Vertical Asymptote: $x = \frac{1 + i \sqrt{2}}{3}, x = \frac{1 - i \sqrt{2}}{3}$

$\begin{array}{c} x & y \\ 1 & 0.693 \\ 2 & 2.197 \\ 3 & 3.091 \end{array}$

Step 6